Suppose we have primal and its dual in standard form, that is
\begin{align*} (P) \max z = cx \\ st \; \; Ax = b \\ \; \; \; x \geq 0 \\ \end{align*}
\begin{align*} (D) \min z = by \\ st \; \; yA \geq c \\ \; \; \; y \; \; \; free \\ \end{align*}
Where $A$ is an $m $ by $n$ matrix an $x$ is an n vector and $y$ is an m vector.
Suppose we multiply one of the constraints of the primal by some number $\alpha > 0$. Does this affect the solution of the dual?
Thoughts:
Since a constraint is of the form $a_{ij} \cdot x $, take one of the $i's$, say we multiply
$$ a_{i1}x_1 + a_{i2} x_2 + ... + a_{in} x_n $$
by $\alpha $
Once we set up our tableau, once we divide this row by $\alpha$, then in the LFH, we would have
$$ \frac{ b_i}{\alpha} $$
the ith component of the vector $b$. Doesnt it change the solution in the primal tableau? Since solutions are the same for primal and dual???
Multiplication by a non-zero scalar is equivalent to multiplication of an elementary matrix, $E$.
\begin{align*} (P') \max z = c^Tx \\ st \; \; (EA)x = (Eb) \\ \; \; \; x \geq 0 \\ \end{align*}
The dual is \begin{align*} (D') \min z = (Eb)^Ty \\ st \; \; y^T(EA) \geq c \\ \; \; \; y \; \; \; free \\ \end{align*}
Suppose $w$ is the original dual solution, then $y=E^{-T}w.$
For the operation of multiplication by a scalar, we have $E^T=E$.
Hence $y=E^{-1}w$. That is if we multiply $\alpha$ to the $i$-th constraint, now for the dual solution, we would divide $w_i$ by $\alpha$ and we can keep the rest to be the same.